Cho (x + y + z)^2 = x^2 + y^2 + z^2. Chứng minh rằng: 1/x^3 + 1/y^3 + 1/3xyz
Cho (x + y + z)2 = x2 + y2 + z2.
Chứng minh rằng: \[\frac{1}{{{x^3}}} + \frac{1}{{{y^3}}} + \frac{1}{{{z^3}}} = \frac{3}{{xyz}}\]
Ta có: (x + y + z)2 = x2 + y2 + z2
Û x2 + y2 + z2 + 2(xy + yz + zx) = x2 + y2 + z2
Û xy + yz + zx = 0
Lại có: \[\frac{1}{{{x^3}}} + \frac{1}{{{y^3}}} + \frac{1}{{{z^3}}} - \frac{3}{{xyz}}\]
\[ = \left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)\left( {\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} - \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}} \right)\]
\[ = \left( {\frac{{xy + yz + zx}}{{xyz}}} \right)\left( {\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} - \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}} \right) = 0\]
Suy ra \[\frac{1}{{{x^3}}} + \frac{1}{{{y^3}}} + \frac{1}{{{z^3}}} = \frac{3}{{xyz}}\](đpcm)
Vậy \[\frac{1}{{{x^3}}} + \frac{1}{{{y^3}}} + \frac{1}{{{z^3}}} = \frac{3}{{xyz}}\].